Difference between revisions of "Equivalent conditions for a linear map between two normed spaces to be continuous everywhere"

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(Created page with "==Statement== Given two normed spaces {{M|(X,\Vert\cdot\Vert_X)}} and {{M|(Y,\Vert\cdot\Vert_Y)}} and a linear map {{M|L:X\rightarrow Y}} between them, th...")
 
m (Adding claim 1) implies 2))
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==Statement==
 
==Statement==
 
Given two [[normed space|normed spaces]] {{M|(X,\Vert\cdot\Vert_X)}} and {{M|(Y,\Vert\cdot\Vert_Y)}} and a [[linear map]] {{M|L:X\rightarrow Y}} between them, then{{rAPIKM}} the following are equivalent (meaning if you have 1 you have all the others):
 
Given two [[normed space|normed spaces]] {{M|(X,\Vert\cdot\Vert_X)}} and {{M|(Y,\Vert\cdot\Vert_Y)}} and a [[linear map]] {{M|L:X\rightarrow Y}} between them, then{{rAPIKM}} the following are equivalent (meaning if you have 1 you have all the others):
−
# If we have a sequence {{M|1=(x_n)_{n=1}^\infty\subseteq X}} with {{M|x_n\rightarrow 0}} then {{M|1=(\Vert L(x_n)\Vert_Y)_{n=1}^\infty}} is a [[bounded sequence]]
+
# If we have a sequence {{M|1=(x_n)_{n=1}^\infty\subseteq X}} with {{M|x_n\rightarrow 0}} then {{M|1=(\Vert L(x_n)\Vert_Y)_{n=1}^\infty}} is a [[bounded sequence]]<ref group="Note">this may be better said (perhaps) as "{{M|L}} maps all sequences convergent to {{M|0}} to [[bounded sequence|bounded sequences]]"</ref>
 
# {{M|L}} is [[continuous]] at a point (any point)
 
# {{M|L}} is [[continuous]] at a point (any point)
 
# {{M|L}} is a [[bounded linear map]], that is {{M|\exists A>0\ \forall x\in X[\Vert L(x)\Vert_Y\le A\Vert x\Vert_X]}}
 
# {{M|L}} is a [[bounded linear map]], that is {{M|\exists A>0\ \forall x\in X[\Vert L(x)\Vert_Y\le A\Vert x\Vert_X]}}
 
# {{M|L}} is [[continuous|continuous everywhere]]
 
# {{M|L}} is [[continuous|continuous everywhere]]
 
==Proof==
 
==Proof==
 +
{{Begin Inline Theorem}}
 +
{{M|1)\implies 2)}}: That {{M|L}} maps all [[null sequence|null sequences]] to [[bounded sequence|bounded sequences]] {{M|\implies}} {{M|L}} is [[continuous]] at a particular point
 +
{{Begin Inline Proof}}
 +
{{:Equivalent conditions for a linear map between two normed spaces to be continuous everywhere/1 implies 2}}
 +
{{End Proof}}{{End Theorem}}
 +
<br/><br/>
 
{{Requires proof|See page 154 in<ref name="APIKM"/>}}
 
{{Requires proof|See page 154 in<ref name="APIKM"/>}}
 +
==Notes==
 +
<references group="Note"/>
 
==References==
 
==References==
 
<references/>
 
<references/>
 
{{Theorem Of|Linear Algebra|Functional Analysis|Metric Space}}
 
{{Theorem Of|Linear Algebra|Functional Analysis|Metric Space}}

Revision as of 18:41, 26 February 2016

Statement

Given two normed spaces [ilmath](X,\Vert\cdot\Vert_X)[/ilmath] and [ilmath](Y,\Vert\cdot\Vert_Y)[/ilmath] and a linear map [ilmath]L:X\rightarrow Y[/ilmath] between them, then[1] the following are equivalent (meaning if you have 1 you have all the others):

  1. If we have a sequence [ilmath](x_n)_{n=1}^\infty\subseteq X[/ilmath] with [ilmath]x_n\rightarrow 0[/ilmath] then [ilmath](\Vert L(x_n)\Vert_Y)_{n=1}^\infty[/ilmath] is a bounded sequence[Note 1]
  2. [ilmath]L[/ilmath] is continuous at a point (any point)
  3. [ilmath]L[/ilmath] is a bounded linear map, that is [ilmath]\exists A>0\ \forall x\in X[\Vert L(x)\Vert_Y\le A\Vert x\Vert_X][/ilmath]
  4. [ilmath]L[/ilmath] is continuous everywhere

Proof

[ilmath]1)\implies 2)[/ilmath]: That [ilmath]L[/ilmath] maps all null sequences to bounded sequences [ilmath]\implies[/ilmath] [ilmath]L[/ilmath] is continuous at a particular point


This is a proof by contrapositive. That is we will show that if [ilmath]L[/ilmath] is not continuous at [ilmath]p[/ilmath] [ilmath]\implies[/ilmath] [ilmath]L[/ilmath] takes a null sequence to one that isn't bounded (an unbounded one).

  • Let the normed spaces [ilmath]X[/ilmath] and [ilmath]Y[/ilmath] be given, as well as a linear map [ilmath]L:X\rightarrow Y[/ilmath]
    • Suppose that [ilmath]L[/ilmath] is not continuous at [ilmath]p[/ilmath], this means:
  • [ilmath]\exists (x_n)_{n=1}^\infty\rightarrow p[/ilmath] such that [math]\lim_{n\rightarrow\infty}\left(L(x_n)\right)\ne L(p)[/math]

Recall that continuity states that:

  • [ilmath]L[/ilmath] is continuous at [ilmath]p[/ilmath] [ilmath]\iff[/ilmath] [math]\forall(x_n)_{n=1}^\infty\subseteq X \left[\left(\lim_{n\rightarrow\infty}(x_n)=p\right)\implies\left(\lim_{n\rightarrow\infty}\left(L(x_n)\right)=L\left(\lim_{n\rightarrow\infty}(x_n)\right)=L(p)\right)\right][/math]

So it follows that to not be continuous at [ilmath]p[/ilmath]:

  • [math]\exists (x_n)_{n=1}^\infty\subseteq X\left[\left(\lim_{n\rightarrow\infty}(x_n)=p\right)\wedge\neg\left(\lim_{n\rightarrow\infty}(L(x_n))=L(p)\right)\right][/math], by negation of implies. Additionally we may negate the [ilmath]=[/ilmath] and thus we see this is the same as:
    • [math]\exists (x_n)_{n=1}^\infty\subseteq X\left[\left(\lim_{n\rightarrow\infty}(x_n)=p\right)\wedge\left(\lim_{n\rightarrow\infty}(L(x_n))\ne L(p)\right)\right][/math]

Which is exactly "there exists a sequence in [ilmath]X[/ilmath] whose limit is [ilmath]p[/ilmath] and where [ilmath]\lim_{n\rightarrow\infty}(L(x_n))\ne L(p)[/ilmath]"

  • Let us now take [ilmath]L(x_n)\not\rightarrow L(p)[/ilmath] and subtract [ilmath]L(p)[/ilmath] from both sides. We see:
    • [ilmath]L(x_n)-L(p)\not\rightarrow L(p)-L(p)[/ilmath], using the fact that [ilmath]L[/ilmath] is linear we see that:
      • [ilmath]L(x_n-p)\not\rightarrow L(0)[/ilmath] and [ilmath]L(0)=0\in Y[/ilmath] so:
    • [ilmath]L(x_n-p)\not\rightarrow 0[/ilmath]
  • Thus [ilmath]\Vert L(x_n-p)\Vert_Y\not\rightarrow 0[/ilmath] (as [ilmath]\Vert0\Vert_Y=0[/ilmath] by definition)
  • So [math]\exists C>0\ \forall N\in\mathbb{N}\ \exists n\in\mathbb{N}[n>N\wedge\Vert L(x_n-p)\Vert_Y>\epsilon][/math]

If a sequence converges to [ilmath]0[/ilmath] then we have:

  • [ilmath]\forall\epsilon>0\ \exists N\in\mathbb{N}\ \forall n\in\mathbb{N}[n>N\implies d(x_n,x)<\epsilon][/ilmath] we know we don't have this, so we negate it:
  • [ilmath]\exists\epsilon>0\ \forall N\in\mathbb{N}\ \exists n\in\mathbb{N}[n>N\wedge d(x_n,x)>\epsilon][/ilmath].

That is there exists an [ilmath]\epsilon>0[/ilmath] such that for all [ilmath]N[/ilmath] there exists a bigger [ilmath]n[/ilmath] such that [ilmath]d(x_n,x)>\epsilon[/ilmath], we shall later call such an [ilmath]\epsilon[/ilmath] [ilmath]C[/ilmath] and construct a subsequence out of the [ilmath]n[/ilmath]s

  • Thus it is possible to construct a subsequence, [ilmath](\Vert L(x_{n_k}-p)\Vert_Y)_{k=1}^\infty[/ilmath] of the image [ilmath](x_n)[/ilmath] where for every [ilmath]k[/ilmath] we have:
  • [ilmath]\Vert L(x_{n_k}-p)\Vert_Y>C[/ilmath]

By the negation of convergent sequence we see that there exists a [ilmath]C[/ilmath] such that for all [ilmath]N\in\mathbb{N} [/ilmath] there exists another natural number, [ilmath]n[/ilmath] such that [ilmath]\Vert L(x_n-p)\Vert_Y>C[/ilmath], we construct a sequence, [ilmath](n_k)_{k=1}^\infty[/ilmath] of such [ilmath]n[/ilmath]-ns. That is we know there exists a [ilmath]C[/ilmath], so we pick [ilmath]N=1[/ilmath] and get the [ilmath]n[/ilmath] that works, this is our first term, we then set [ilmath]N=2[/ilmath] and the [ilmath]n[/ilmath] that works is our second term, and so forth.

Some of these [ilmath]n[/ilmath]s may be the same, but that doesn't matter.

[ilmath](x_{n_k})[/ilmath] is the subsequence of [ilmath](x_n)[/ilmath] which contains only the terms that satisfy [ilmath]\Vert L(x_n-p)\Vert_Y> C[/ilmath]

It would have been better if I used [ilmath]n[/ilmath] and [ilmath]m[/ilmath] as terms, to make which [ilmath]n[/ilmath] I am talking about clearer, but a reader able to attempt this proof should follow.

  • We now have a sequence [ilmath](x_{n_k})[/ilmath] such that [ilmath]\Vert L(x_{n_k}-p)\Vert_Y>C[/ilmath]
  • Define a new sequence [math]b_k:=\frac{1}{\sqrt{\Vert x_{n_k}-p\Vert} }[/math]
    • It is easy to see that [ilmath]b_k\rightarrow +\infty[/ilmath] (as [ilmath](x_{n_k}-p)\rightarrow 0[/ilmath])

TODO: Prove that this tends to [ilmath]+\infty[/ilmath]

  • Define a new sequence [math]d_k:=b_k(x_{n_k}-p)[/math]
  • Clearly [math]d_k\rightarrow 0[/math]

If [ilmath](d_k)_{k=1}^\infty\rightarrow 0[/ilmath] then [ilmath]\forall\epsilon>0\exists N\in\mathbb{N}\forall n\in\mathbb{N}[n>N\implies\Vert d_k-0\Vert_X<\epsilon][/ilmath].

  • Notice [math]\Vert d_k\Vert_X=b_k\Vert x_{n_k}-p\Vert_X=\frac{\Vert x_{n_k}-p\Vert_X}{\sqrt{\Vert x_{n_k}-p\Vert_X} }[/math] [math]=\sqrt{\Vert x_{n_k}-p\Vert_X}[/math]

TODO: Finish this proof

  • But [math]\Vert L(d_k)\Vert_Y=\Vert L(b_k(x_{n_k}-p))\Vert_Y=b_k\Vert L(x_{n_k}-p)\Vert_Y\ge Cb_k\rightarrow +\infty[/math]
  • Thus we have shown if [ilmath]L[/ilmath] is not continuous at [ilmath]p[/ilmath] that the mapping of a null sequence is unbounded, the contrapositive of what we set out to claim



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See page 154 in[1]

Notes

  1. ↑ this may be better said (perhaps) as "[ilmath]L[/ilmath] maps all sequences convergent to [ilmath]0[/ilmath] to bounded sequences"

References

  1. ↑ 1.0 1.1 Analysis - Part 1: Elements - Krzysztof Maurin


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